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What is the orbital velocity and period of a ring particle at the outer edge of Saturn’s A ring? (Hint: Use the formula for circular velocity, Eq. 5-1a. The formula requires input quantities in kg and m.) (Note: The radius of the outer edge of the A ring is 136,500 km.)
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The orbital velocity and period of a ring particle at the outer edge of Saturn’s A ring.
Answer to Problem 9P
The orbital velocity of a ring particle at the outer edge of Saturn’s A ring is 1.67×104 m/s and orbit period of the ring particle is 14.3 h.
Explanation of Solution
Write the expression for the orbital velocity.
Vs=√GMsRs (I)
Here, Vs is the orbital velocity of Saturn’s A ring, G is the universal gravitational constant, Ms is the mass of Saturn and Rs is the radius of the edge of A ring.
Write the expression for the time period.
Ts=2πRsVs (II)
Here, Ts is the time period of a ring particle at the outer edge of Saturn’s A ring.
From the celestial profile, the mass of the Saturn is 5.68×1026 kg.
Conclusion:
Substitute 6.67×10−11 N⋅m2/kg2 for G, 5.68×1026 kg for Ms and 136500 km for Rs in equation (I) to find Vs.
Vs=√(6.67×10−11 N⋅m2/kg2)(5.68×1026 kg)(136500 km)=√(6.67×10−11 N⋅m2/kg2)(5.68×1026 kg)(136500 km)(103 m1 km)=√(6.67×10−11 N⋅m2/kg2)(5.68×1026 kg)(136500×103 m)=1.67×104 m/s
Substitute 136500 km for Rs and 1.67×104 m/s for VS in equation (II) to find Ts.
Ts=2π(136500 km)(1.67×104 m/s)=2π(136500 km)(103 m1 km)(1.67×104 m/s)=(5.14×104 s)(1 h3600 s)=14.3 h
Therefore, the orbital velocity of a ring particle at the outer edge of Saturn’s A ring is 1.67×104 m/s and orbit period of the ring particle is 14.3 h.
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